"Spread operator" syntax

I have this base type:

export interface IMyType : {
   field1: number,
   field2: number
}

I also have this "extension" type:

export interface ISelectable {
   isSelected: boolean;
}

Somewhere else, I have a function like this:

const foo = (param: (IMyType & ISelectable)[]) => { // <-- my on-the-fly "extended" type
   param[0].isSelected = ... //do whatever
}

My problem is to created the composite array:

const dataArray: IMyType[] = [{field1: 0, field2: 0}, {field1: 0, field2: 0}];
const extendedArray = [ dataArray, ...[{isSelected: false}] ]; // <-- extend!

foo(extendedArray); // <-- PROBLEM! typescript still sees extendedArray
                    //     as of type IMyType[] instead of 
                    //    (IMyType & Iselectable)[]  so it refuses this.

I’m 99% sure that I misused the spread operator (...) but right now I can’t find the right syntax.

In recap, I tried the following :

const extendedArray = [ dataArray, ...[{isSelected: false}] ]; //Nope! (Makes zero sense, in the light of the comments below)
const extendedArray = [ ...dataArray, {isSelected: false} ]; //nope! (Makes zero sense, in the light of the comments below)
const extendedarray = dataArray.map (x => {isSelected: false, ...x}) //Almost correct but was missing sone syntactic sugar!

I’ve read this and many other answers but I can’t figure it out : TypeScript add object to array with spread

>Solution :

Bu using the spread operateor you appened elements, you do not modify the existing ones

So
const extendedArray = [ dataArray, ...[{isSelected: false}] ];

The extendedArray will have two type elements:
First element will be an array of IMyType , the second element will be of type { isSelected: boolean }

If you want to modify all elements in an array – use .map(). And preferably create a copy of all objects like this:

const extendedArray = dataArray.map(item => { return { ...item, isSelected: false } });

//All is ok now
foo(extendedArray);

Playground here